Question

What is the angle between the following vectors?

$\overrightarrow{A}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{B}=-2\hat{i}-2\hat{j}-2\hat{k}$.

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. $0^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is D

${180}^{\circ }$

$\overrightarrow{A}.\overrightarrow{B}=|A||B|cos\theta or,cos\theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|A||B|}$ ----------------(i)

But, $\overrightarrow{A}.\overrightarrow{B}=(\hat{i}+\hat{j}+\hat{k}).(-2\hat{i}-2\hat{j}-2\hat{k})$

$\overrightarrow{A}.\overrightarrow{B}=-2-2-2$

$\overrightarrow{A}.\overrightarrow{B}=-2-2-2=-6$

Again A = $|\overrightarrow{A}|=\sqrt{(1{)}^2+(1{)}^2+(1{)}^2}=\sqrt{3}$;

B = $|\overrightarrow{B}|=\sqrt{(-2{)}^2+(-2{)}^2+(-2{)}^2}=\sqrt{12}=2\sqrt{3}$

Now,$cos\theta =\frac{-6}{\sqrt{3}\times 2\sqrt{3}}=-1\Rightarrow \theta ={180}^{\circ }$